机原作业答案(正确的).doc

REFERENCE ANSWERforEXERCISE BOOKofMechanisms and Machine TheoryTRG of Machinery Theory and DesignCollege of Mechanical EngineeringFuzhou University2003Name Class Student No. Date 2-1 Draw the kinematic diagrams of the mechanisms shown in Fig2-1.Fig2-1(b)Fig2-1(a)Name Class Student No. Date 2-2 Draw the kinematic diagrams of the mechanisms shown in Fig2-2.Fig2-2(a)Name Class Student No. Date 2-3 Draw the kinematic diagram of the mechanism shown in Fig2-3.Fig2-3Name Class Student No. Date Fig2-4(a)2-4 Calculate the degree of freedom of the mechanisms shown in Fig2-4,and point out what should be paid attention to during the calculation .解：(a) 左右为对称结构，设左侧为虚约束。(b) E为杆4、5、6的复合铰链。(c) 滑块7与机架8间为移动副。F=3n-2PL-Ph=37-210=1Fig2-4(b)解： (1)红线内的构件为重复结构，构成虚约束。(2)去掉以上构件后，C仍为构件2、3、4的复合铰链。(3)滑块5与机架6之间为移动副。F= 3n-2PL-Ph =3527=1Name Class Student No. Date 2-5 Calculate the degree of freedom of the mechanisms shown in Fig2-5,and point out what should be paid attention to during the calculation .Fig2-5(a)解：(a) 两个滚子有局部自由度。(b) 滚子D与凸轮1之间只能算一个高副。F=3n-2PL-Ph =37 29 -2=1Fig2-5(b)解：(1) 杆件BC与齿轮2焊接在一起。 (2) A 为齿轮4、杆件1和机架5的复合铰链。F= 3n-2PL-Ph =34 25 -1 =1常见错误：认为B是复合铰链，而不认为A是复合铰链。Name Class Student No. Date 2-6 Calculate the degree of freedom of the mechanisms shown in Fig2-6,and point out what should be paid attention to during the calculation .Fig2-6(a)解：(a) C为构件2、3、4的复合铰链。(b) C处有两个转动副和两个移动副。(c) E处有一个转动副和两个移动副。F= 3n-2PL-Ph =37 210=1注意：E不是复合铰链！ Fig2-6(b)解：当构件尺寸任意时，构件2作平面复杂运动，而杆4与机架间组成移动副，所以杆4仅作平动。因此，构件2和构件4之间有相对转动。因此，应该有构件6，并且构件4和6之间有转动副，如右图所示。当ABCD 且BCAD时，杆2仅作平动。杆4与机架间组成移动副，所以杆4也仅作平动。这样，构件2和构件4之间就没有相对转动，只有相对移动。即：构件4和构件6之间就没有相对转动了，因此，可将构件6与构件4焊接起来（去掉构件6），如左图所示。然而，在计算机构自由度时，应该按一般尺寸情况下进行分析，即：应该按照右图情况来分析机构的自由度。F= 3n-2PL-Ph =35 27=1Name Class Student No. Date 2-7 The kinematic diagram of an engine mechanism is given in Fig2-7.(1) Calculate the degree of freedom of the mechanism, and point out what should be paid attention to during the calculation.(2) Make the structural analysis for the mechanism.(3) Make the structural analysis for the mechanism when link EFG is regarded as the driver.Note: During structural analysis, list the assembly order of Assur groups, the type of group, the grade of group, the grade of the mechanism, the link serial numbers, the inner pair and the outer pairs of each group in each mechanism.解：（1） F= 3n-2PL-Ph =37 210=1（2） 当AB为原动件时，（3）当EFG为原动件时，Name Class Student No. Date 2-8 Make the structural analysis for the mechanism shown in Fig2-8.(a) When link 1 is regarded as the driver. (b) When link 5 is regarded as the driver.Note: During structural analysis, list the assembly order of Assur groups, the type of group, the grade of group, the grade of the mechanism, the link serial numbers, the inner pair and the outer pairs of each group in each mechanism.Fig2-8解：(a) 当1为原动件时 杆件 2, 3, 4 和 5组成一个三级 Assur group.(b) 当5为原动件时杆件3和4组成第一个 RPR Assur group.杆件1和2组成第二个 RPR Assur groupName Class Student No. Date 2-9 The schematic diagram of a punch machine designed by someone is shown in Fig2-9. This machine should be able to transform a continuous rotation of gear 1 into a translation of the punch 4. Can the machine work properly? If it cant ,please rectify it.Fig2-9解：不能正常工作。改正如图(或者改成题目2-3构件5、6、7的连接)Name Class Student No. Date 2-10 The schematic diagram of a mechanism designed by someone is shown in Fig2-10. This mechanism should be able to transform a continuous rotation of link1 into an oscillation of link4. Can the mechanism work properly? If it cant ,please rectify it.Fig2-10解：不能正常工作。改正后 Name Class Student No. Date Fig3-1(a)3-1 Locate all instant centres of mechanisms in the position shown in Fig3-1.Fig3-1(c)Fig3-1(b)Fig3-1(d)Fig3-1(e)Name Class Student No. Date Fig3-1(f)Name Class Student No. Date 3-2 In the position shown in Fig3-2, determine the ratio w3/w1 of the angular velocity of gear 3 to that of gear 1, using the method of instant centres.Fig3-2解：P13是构件1和3的瞬心，等速重合点，所以w1LAE =w3LDE w3/w1=LAE/LDE3-3 In the position shown in Fig3-3, determine the ratio w2/w1 of the angular velocity of follower 2 to that of cam 1, using the method of instant centres.Fig3-3解：E（P12）是构件1和2的瞬心，等速重合点，所以w1LOE=w2LAEw2/w1=LOE/LAEName Class Student No. Date 3-4 In the pivot four-bar linkage shown in Fig3-4, w1= 10rad/sec. Using the method of instant centres, (a) Find the velocity of point C in the position shown in the figure.(b) In the position shown in the figure, locate the point E on the line BC (or its extension) which has the minimum velocity among all points of line BC and its extension, and then calculate its velocity.Fig3-4(c) draw two positions of the crank AB when VC=0.解：（a）VB1= w1LAB=VB2= w2LFB ，所以VC= VC2= w2LFC= w1LABLFC/LFB (b)VE=w2LFE。(c ) VC=0所对应的曲柄AB的两个位置:Name Class Student No. Date 3-5 In the six-bar mechanism shown in Fig3-5, XA=0, YA=0, XD=450mm, YD=0, LAB=150mm, LBC=400mm, LDC=350mm, CDE=30, LDE=150mm, LEF=400mm. The crank AB rotates at a constant speed 10rad/sec.Write a main program to analyze the output motion of the point F when the driver AB rotates from 0 to 360 with a step size of 5. Fig3-5解：FOR I=0 TO 360 STEP 5 CALL LINK(0, 0, 0, 0, 0, 0, I*PI/180, 10, 0, 150, XB, YB, VBX, VBY, ABX, ABY) CALL RRR(450, 0, 0, 0, 0, 0, XB, YB, VBX,VBY, ABX,ABY,350,400,Q3,W3,E3,Q2,W2,E2) CALL LINK(450, 0, 0, 0, 0, 0, Q3-PI/6, W3, E3, 150, XE, YE, VEX, VEY, AEX, AEY) CALL RRP(1, 0, 400, XE, YE, VEX, VEY, AEX, AEY, Q5, W5, E5) CALL LINK(XE,YE,VEX,VEY,AEX,AEY,Q5,W5, E5, 400, XF, YF, VFX, VFY, AFX, AFY) PRINT I, XF, YF, VFX, VFY, AFX, AFYNEXT IName Class Student No. Date 3-6 In the mechanism shown in Fig3-6,XA=0, YA=0, XD=200mm, YD=0, LAB=80mm, LCD=60mm and LBE=380mm. The crank AB rotates at a constant speed of 10rad/sec. Write a main program to analyze the output motion of the point E when the driver AB rotates from 0 to 360 with a step size of 5.Fig3-6FOR I=0 TO 360 STEP 5 CALL LINK(0, 0, 0, 0, 0, 0, I*PI/180, 10, 0, 80, XB, YB, VBX, VBY, ABX, ABY) CALL RPR(1, XB, YB, VBX, VBY, ABX, ABY, 200, 0, 0, 0, 0, 0, 60, Q2, W2, E2) CALL LINK(XB, YB, VBX, VBY, ABX, ABY, Q2, W2, E2, 380, XE, YE, VEX, VEY, AEX, AEY) PRINT I, XE, YE, VEX, VEY, AEX, AEYNEXT IName Class Student No. Date Fig3-73-7 . In the mechanism shown in Fig3-7, XG=YG=0, XB= - 42, YB=39, XD=70, YD=75, LBA=34mm, LGF=24mm, LFE=95mm, LEC=69mm, LDC=48mm, EFG=90. The crank BA rotates at a constant speed of 10 rad/sec. Write a main program to analyze the output motion of the point C when the driver BA rotates from 0 to 360 with a step size of 5.FOR I=0 TO 360 STEP 5 CALL LINK(-42, 39, 0, 0, 0, 0, I*PI/180, 10, 0, 34, XA, YA, VAX, VAY, AAX, AAY) CALL RPR(-1, 0, 0, 0, 0, 0, 0, XA, YA, VAX, VAY, AAX, AAY, 24, QFE, W3, E3) CALL LINK(0, 0, 0, 0, 0, 0, QFE+PI/2, W3, E3, 24, XF, YF, VFX, VFY, AFX, AFY) CALL LINK(XF, YF, VFX, VFY, AFX, AFY, QFE, W3, E3,95, XE, YE, VEX, VEY, AEX, AEY) CALL RRR(XE, YE, VEX, VEY, AEX, AEY, 70, 75, 0, 0, 0, 0, 69, 48,QEC, W4, E4, QDC, W5, E5) CALL LINK(70, 75, 0, 0, 0, 0, QDC, W5, E5, 48, XC, YC, VCX, VCY, ACX, ACY) PRINT I, XC, YC, VCX, VCY, ACX, ACYNEXT IName Class Student No. Date 3-8 In the six-bar mechanism shown in Fig3-8, XB=0, YB=0, XF=37.2, YF=17.5, YC=28.8, LFE=16.8mm, LEC=39.2mm, LCD=20.633mm, LDE=36.4mm, BGA=90, LBG=9mm, LGA=58mm. The crank FE rotates clockwise at a constant speed of 10 rad/sec. Write a main program to analyze the output motion of the point A.when the driver FE rotates from 360 to 0 with a step size of -5.Fig3-8For I = 360 To 0 Step -5Call LINK(37.2, 17.5, 0, 0, 0, 0, 16.8, I*PI/180, 10, 0, XE, YE, VEX, VEY, AEX, AEY)Call RRP(-1, 28.8, XE, YE, VEX, VEY, AEX, AEY, 39.2, QEC, W4, E4, XC)Call LINK(XE, YE, VEX, VEY, AEX, AEY, QEC-QCED, W4,E4,XD,YD, VDX, VDY, ADX, ADY)Call RPR(1, 0, 0, 0, 0, 0, 0, XD, YD, VDX, VDY, ADX, ADY, LBG, QGA, W2, E2)Call LINK(0, 0, 0, 0, 0, 0, LBG, 3 * PI / 2 + QGA, W2, E2, XG, YG, VGX, VGY, AGX, AGY)Call LINK(XG, YG, VGX, VGY, AGX, AGY, LGA, QGA, W2, E2, XA, YA, VAX, VAY, AAX, AAY)Print I, XA, YA, VAX, VAY, AAX, AAYNext IName Class Student No. Date Fig4-1(b)4-1 Determine the type of the pivot four-bar linkages whose dimensions are shown in Fig4-1.Fig4-1(a)双曲柄机构 曲柄摇杆机构双摇杆机构 双摇杆机构Fig4-1(d)Fig4-1(c)Name Class Student No. Date 4-2 In the revolute four-bar mechanism shown in Fig4-2(1) Find the pressure angle a and the transmission angle g of the mechanism in the position shown in the figure.(2) Find the angular stroke ymax of the link DC.(3) Find the crank acute angle q between the two limiting positions.(4) Find the maximum pressure angle amax and the minimum transmission angle gmin.(5) Is there dead-point during the whole cycle of the motion when the link DC is regarded as the driver? If there is, when? And draw the dead-point positions of the mechanism.Fig4-2(5)当DC为原动件时，此机构有死点位置。死点位置为图中AB1C1D和AB2C2D。Name Class Student No. Date 4-3 Design an offset slider-crank mechanism ABC in which the crank AB is the driver. The maximum pressure angle amax=30. Find the stroke H of the slider and the crank acute angle q between the two limiting positions.Fig4-34-4 Determine the angular strokes of the rockers AB and CD shown in Fig4-4, respectively, using graphical method.Name Class Student No. Date 4-5 In Fig4-5 there are two positions, B1C1 and B2C2, of the coupler BC of a revolute four-bar linkage ABCD. The link AB is a driver. The pressure angle a at the first position is 0o. The second position of the mechanism is a toggle position. Design the linkage and write out the drawing steps briefly.解：a) 链点A 必在B1B2的垂直平分线上。 b) 类似的，铰链点D必在C1C2的垂直平分线上。c) AB为原动件时，机构在第一位置的压力角为 0o 得到B1C1C1D。d) 机构的第二位置为一死点位置得到A, B2, C2三点共线。Name Class Student No. Date 4-6 In a crank-slider mechanism, two sets of corresponding positions of the slider and a line segment AE on the crank ABE are shown in Fig4-6. The position C1 of the slider is its left limiting position. Find the first position B1 of the revolute B and write out the drawing steps briefly.Fig4-6解：作D AC2E1DAC2E2，且字母旋向相同，得C2 。因C1为滑块的极限位置之一，所以B1点在AC1连线上。作C1C2的中垂线与AC1交于待定活动铰链点B的第一个位置点B1。Name Class Student No. Date 4-7 In a revolute four-bar linkage ABCD, side link AB is the driver. Two sets of corresponding positions of the side link CD and a line segment CE on the coupler CBE are shown in Fig4-7. The first position of the linkage is also a dead point. Find the second position B2 of the revolute B and write out the drawing steps briefly.Fig4-7解：作D A2C1E1DAC2E2，得A2点。因AB为原动件且机构第一位置为死点，所以B1点在DC1的延长线上。作AA2的中垂线与DC1的延长线交于待定活动铰链点B的第一个位置点B1。Name Class Student No. Date 4-8 In a crank-rocker linkage ABCD, side link AB is the driver. Two positions of the rocker CD are shown in Fig4-8. At the first position , the pressure angle of the linkage is zero. Position DC2 is one of the limit positions of the rocker. Find the first position B1 of the revolute B and write out the drawing steps briefly.Fig4-8解：因A点在C2B2延长线上, 故在C1B1延长线上截取C1A2=C2A，得A2点。作AA2的中垂线与C1B1交于待定活动铰链点B的第一个位置点B1。Name Class Student No. Date 4-9 In an offset slider-crank mechanism ABC, two sets of corresponding positions of the crank AB and a point F on the slider are shown in Fig4-9. When the crank AB is located at position AB1, the slider reaches its left limit position. Find the first position C1 of the revolute C on the slider and write out the drawing steps briefly.Fig4-9解：作B2B2 /=F1F2，得B2点因为第一位置是滑块的左极限位置,故C1在B1A的延长线上. 作B1B2的中垂线，交B1A的延长线于待定活动铰链点C的第一个位置点C1。作C1C2=F1F2，得C2点。Name Class Student No. Date 4-10 In a crank-rocker mechanism ABCD, the coefficient of travel speed variation(k) is1.25, when crank AB rotate constantly. The angular stroke of rocker DC y=60o .The length of the rocker DC is 40mm. The length of the frame AD is 50mm. Design the mechanism and write out the drawing steps briefly.解：（取比例尺l =0.001m/mm）取DC1=DC2=40mm，作直角三角形C1C2H,其中C1C2H=90=70 作三角形C1C2H的外接圆，以D点为圆心，以50mm为半径作圆，两圆的交点为铰链点A。连接AC1、AC2,以A点为圆心，以（AC2 AC1）2为半径作圆得铰链点B,AB=17.98mmBC=53.72mmName Class Student No. Date 4-11 In an offset slider-crank mechanism, the offset e is 20mm. The coefficient of travel speed variation k is 1.3. The working stroke H of the slider is 50mm. Design the offset slider-crank mechanism解： （取比例尺l =0.001m/mm）作C1C2=50mm,作等腰三角形OC1C2其中OC1C2=OC2C1=90=86.5,以O为圆心，以OC1为半径作圆作一与C1C2平行且距离为20mm的直线，此直线与圆O的交点为铰链点A以A为圆心，以（AC2AC1）2为半径作圆A得到铰链点B。BC=55.06mm,AB=22.83mm.Name Class Student No. Date 5-1 A plate cam with positive-offset translating roller follower has the following motion: a rise through lift h=40mm with a sine acceleration motion curve during d0=150, dS=30, a return with a 3-4-5 polynomial motion curve during d0=120, and dS=60. The cam rotates clockwise. And rP=40mm. rR=12mm, e=12mm and rC=25mm. Construct the pitch curve and the cam contour graphically with a scale of 1:1. Label in red the centerline of the follower, S, the roller and a when d = 60 and d = 0.d0306090120150180210240270300330360S(mm)07.8615.7223.5831.44404035.86204.14000Name Class Student No. Date 5-2 A plate cam with translating offset roller follower has the same motion curve and dimensions as exercise5-1 has. Write a program to calculate the co-ordinates of the pitch curve, the cam contour and the locus of the center of the milling cutter, a and rB.SET WINDOW -100,120,-100,60LET M=-1LET N=+1LET RP=40LET E=12LET RR=12LET RC=25LET H=40LET DELTA0=150*PI/180LET DELTAS=30*PI/180LET DELTA01=120*PI/180LET DELTAS1=60*PI/180LET S0=SQR(RP2-E2)BOX CIRCLE -RP,RP,-RP,RPBOX CIRCLE -E,E,-E,EFOR I=0 TO 360 STEP .005 LET DELTA=I*PI/180 IF DELTA=DELTA0 THEN LET DD=DELTA/DELTA0 LET S=H*(DD-1/2/PI*SIN(2*PI*DD) LET S1=H/DELTA0*(1-COS(2*PI*DD) LET S11=2*PI*H/DELTA02*SIN(2*PI*DD) ELSEIF DELTA=(DELTA0+DELTAS) THEN LET S=H LET S1=0 LET S11=0 ELSEIF DELTA=(DELTA0+DELTAS+DELTA01) THEN LET D4=(DELTA-DELTA0-DELTAS)/DELTA01 LET S=H*(1-10*D43+15*D44-6*D45) LET S1=-H/DELTA01*(30*D42-60*D43+30*D44) LET S11=-H/DELTA012*(60*D4-180*D42+120*D43) ELSE LET S=0 LET S1=0 LET S11=0 END IF LET XB=M*(S0+S)*SIN(DELTA)+N*E*COS(DELTA) LET YB=(S0+S)*COS(DELTA)-N*E*SIN(DELTA) LET XB1=M*(S1*SIN(DELTA)+(S0+S)*COS(DELTA)-N*E*SIN(DELTA) LET YB1=S1*COS(DELTA)-(S0+S)*SIN(DELTA)-N*E*COS(DELTA) LET KB=SQR(XB12+YB12) LET XT=XB+M*RR*YB1/KB LET YT=YB-M*RR*XB1/KB LET XC=XT-M*RC*YB1/KB LET YC=YT+M*RC*XB1/KB LET ALAFA=ARCTAN(ABS(S1-N*E)/(S0+S) LET LOUB=(S0+S)2+(S1-N*E)2)1.5/(-(S0+S)*(S11-S0-S)+(S1-N*E)*(2*S1-N*E) PLOT XB,YB PLOT XT,YT PLOT XC,YCNEXT IENDName Class Student No. Date 5-3 In the plate cam with translating offset roller follower as shown in Fig5-1, GH and IJ are two arcs with center at O. Indicate radius of prime circle rP, offset e, cam angle for rise d0, cam angle for outer dwell dS, cam angle for return d0, cam angle for inner dwell dS and lift h. In the position shown in Fig5-1, indicate pressure angle a, displacement S and the corresponding cam angle d.Fig5-1Name Class Student No. Date Fig5-25-4 In the plate cam with translating offset roller follower as shown in Fig5-2, EA, AB and BCD are three arcs with center at O, N and P, respectively. Indicate radius of prime circle rP, offset e, cam angle for rise d0, cam angle for outer dwell dS, ca